uva 10271 dp

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5050;const int maxk = 1010;const int INF = 0x3f3f3f3f;int T, k, n, init[maxn], dp[maxn][maxk];int main(int argc, char const *argv[]){scanf("%d", &T);while (T--){scanf("%d%d", &k, &n);k += 8;for (int i = n; i >= 1; i--)scanf("%d", &init[i]);memset(dp, INF, sizeof(dp));for (int i = 0; i < n; i++)dp[i][0] = 0;for (int i = 3; i <= n; i++)for (int j = 1; j <= k; j++)if (i >= j * 3 && dp[i - 2][j - 1] != INF)dp[i][j] = min(dp[i - 1][j], dp[i - 2][j - 1] + (init[i] - init[i - 1]) * (init[i] - init[i - 1]));printf("%d\n", dp[n][k]);}return 0;}

简单的二维dp，有n个数据，给定k，要从中选出k+8个三元组（x，y，z，其中x<=y<=z)，每选一次的代价为（x-y）^2，求最小代价和。

二维，i表示第i个，j表示前j组，选或者不选两种情况，答案保存在dp[N][K];